These are chat archives for gxyd/sum_convergence

12th
Oct 2015
Aaron Meurer
@asmeurer
Oct 12 2015 01:06
checking products sounds tricky because we’d need to be able to check that the terms are nonzero
although perhaps this is just as hard as checking that the terms are not infinite, which is also a problem for sums
Kalevi Suominen
@jksuom
Oct 12 2015 07:54

I think we could divide the study of infinite sums and products in two parts.

One part is the question whether all terms are well defined, i.e., they are finite in a sum and also non-zero in a product. (Zero is the analogy of (minus) infinity in products as exp(-oo) = 0.)

The second part is the question of convergence after infinities, and zeros in products, have been omitted assuming that their number is finite. This means that we only consider the tail of the sum or product, starting from some point after which all terms are well defined.

This is essentially what the current is_convergent does. For example, in a sum of the form 1/(n2+an+b)\sum 1/(n^2 + an + b), where aa and bb are some constants, the routine will return True even if there may infinities in the term sequence (at most two). (An analogous product would be exp(1/(n2+an+b))\prod \exp(1/(n^2 + an + b)).)

This is how I would like to interpret "convergence"; it should only be concerned with what happens at the limit.

Finding the bad terms is another, independent, matter. There could be a routine for that (finding the zeros of the denominator) but I think it need not be included in is_convergent. It could be the user's responsibility to see that the sum or product is well defined.

Gaurav Dhingra
@gxyd
Oct 12 2015 12:18

Finding the bad terms is another, independent, matter. There could be a routine for that (finding the zeros of the denominator) but I think it need not be included in is_convergent. It could be the user's responsibility to see that the sum or product is well defined.

+1 for this point. Exactly what i have been thinking for the is_convergent().

Gaurav Dhingra
@gxyd
Oct 12 2015 12:33
One more point in favour of @jksuom 's point is: Even the expression is changed(in summation) while we find the convergence, while the index from 1to1 to \infty remains the same. Hence the singularities of the expression are also changed. So including the singularities thing in it, would lead to some expressions being returned as divergent even if it is convergent, even if it contains no singularities in the original expression.
As far as i understood @jksuom 's point. I think the PR looks complete to me. Any thoughts?
Aaron Meurer
@asmeurer
Oct 12 2015 19:54
I would at least mention these caveats in the docstring